JEE Main 2025 Jan 23 Shift 1, Physics Q17: Thin Lens and Mirrors

JEE Main2025Jan 23, Shift 1Physics

Given a thin convex lens (refractive index $μ_{2}$ ), kept in a liquid (refractive index $μ_{1}$ , $μ_{1} < μ_{2}$ ) having radii of curvature $∣ R_{1} ∣$ and $∣ R_{2} ∣$ . Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?

A
$\frac{μ _{1} ∣ R _{1} ∣ \cdot ∣ R _{2} ∣}{μ _{2} ( ∣ R _{1} ∣ + ∣ R _{2} ∣ ) - μ _{1} ∣ R _{1} ∣}$
B
$\frac{μ _{1} ∣ R _{1} ∣ \cdot ∣ R _{2} ∣}{μ _{2} ( ∣ R _{1} ∣ + ∣ R _{2} ∣ ) - μ _{1} ∣ R _{2} ∣}$
C
$\frac{μ _{1} ∣ R _{1} ∣ \cdot ∣ R _{2} ∣}{μ _{2} ( 2∣ R _{1} ∣ + ∣ R _{2} ∣ ) - μ _{1} ∣ R _{1} ∣ \cdot ∣ R _{2} ∣}$
D
$\frac{( μ _{2} + μ _{1} ) ∣ R _{1} ∣}{( μ _{2} - μ _{1} )}$

Solution

A silvered lens acts as an equivalent mirror with power $P_{eq} = 2 P_{lens} + P_{mirror}$ , i.e. $\frac{1}{f _{eq}} = \frac{2}{f _{L}} - \frac{1}{f _{m}}$ , where the back surface acts as a concave mirror of $f_{m} = - ∣ R_{2} ∣/2$ .

Lensmaker's formula in the surrounding liquid: $\frac{1}{f _{L}} = (\frac{μ _{2}}{μ _{1}} - 1) (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ (using the convex/silvered geometry, $R_{1}, R_{2}$ taken with appropriate signs giving the sum $\frac{R _{1} + R _{2}}{R _{1} R _{2}}$ ).

Combining:

$\frac{1}{f _{eq}} = 2 \cdot \frac{μ _{2} - μ _{1}}{μ _{1}} \cdot \frac{R _{1} + R _{2}}{R _{1} R _{2}} + \frac{2}{R _{2}} = \frac{2}{μ _{1} R _{1} R _{2}} [μ_{2} (R_{1} + R_{2}) - μ_{1} R_{2}]$ .

For a real, inverted, same-size image to form at the object location, the object must be at the centre of curvature of the equivalent mirror, i.e. $u = 2 f_{eq}$ .

$u = 2 f_{eq} = \frac{μ _{1} R _{1} R _{2}}{μ _{2} ( R _{1} + R _{2} ) - μ _{1} R _{2}}$ .

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JEE Main 2025 Jan 23 Shift 1, Physics Q17: Thin Lens and Mirrors

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